Integrand size = 26, antiderivative size = 122 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {7 (1-2 x)^{3/2} \sqrt {3+5 x}}{6 (2+3 x)^2}+\frac {637 \sqrt {1-2 x} \sqrt {3+5 x}}{36 (2+3 x)}-\frac {8}{27} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {3035}{108} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]
-8/135*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-3035/108*arctan(1/7*(1 -2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+7/6*(1-2*x)^(3/2)*(3+5*x)^(1/2) /(2+3*x)^2+637/36*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {1}{540} \left (\frac {105 \sqrt {1-2 x} \sqrt {3+5 x} (188+261 x)}{(2+3 x)^2}+32 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-15175 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \]
((105*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(188 + 261*x))/(2 + 3*x)^2 + 32*Sqrt[10] *ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 15175*Sqrt[7]*ArcTan[Sqrt[1 - 2*x ]/(Sqrt[7]*Sqrt[3 + 5*x])])/540
Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {109, 27, 166, 27, 175, 64, 104, 217, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2)^3 \sqrt {5 x+3}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {1}{6} \int \frac {\sqrt {1-2 x} (16 x+223)}{2 (3 x+2)^2 \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{12} \int \frac {\sqrt {1-2 x} (16 x+223)}{(3 x+2)^2 \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{12} \left (\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}-\frac {1}{3} \int -\frac {7039-64 x}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \int \frac {7039-64 x}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {21245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {64}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {21245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {42490}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {6070}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (-\frac {64}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {6070}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\) |
(7*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(6*(2 + 3*x)^2) + ((637*Sqrt[1 - 2*x]*Sq rt[3 + 5*x])/(3*(2 + 3*x)) + ((-64*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5* x]])/3 - (6070*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/6 )/12
3.25.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Time = 1.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.09
| method | result | size |
| risch | \(-\frac {7 \left (-1+2 x \right ) \sqrt {3+5 x}\, \left (188+261 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{36 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (\frac {4 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{135}-\frac {3035 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{216}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(133\) |
| default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (136575 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-288 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+182100 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -384 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +60700 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-128 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+54810 x \sqrt {-10 x^{2}-x +3}+39480 \sqrt {-10 x^{2}-x +3}\right )}{1080 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\) | \(191\) |
-7/36*(-1+2*x)*(3+5*x)^(1/2)*(188+261*x)/(2+3*x)^2/(-(-1+2*x)*(3+5*x))^(1/ 2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)-(4/135*10^(1/2)*arcsin(20/11*x+1/ 11)-3035/216*7^(1/2)*arctan(9/14*(20/3+37/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+1 11*x)^(1/2)))*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {32 \, \sqrt {5} \sqrt {2} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 15175 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 210 \, {\left (261 \, x + 188\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1080 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]
1/1080*(32*sqrt(5)*sqrt(2)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(5)*sqrt(2)* (20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 15175*sqrt(7)* (9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 210*(261*x + 188)*sqrt(5*x + 3)*sqrt(-2*x + 1))/ (9*x^2 + 12*x + 4)
\[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}}}{\left (3 x + 2\right )^{3} \sqrt {5 x + 3}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {4}{135} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3035}{216} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {203 \, \sqrt {-10 \, x^{2} - x + 3}}{12 \, {\left (3 \, x + 2\right )}} \]
-4/135*sqrt(10)*arcsin(20/11*x + 1/11) + 3035/216*sqrt(7)*arcsin(37/11*x/a bs(3*x + 2) + 20/11/abs(3*x + 2)) + 49/18*sqrt(-10*x^2 - x + 3)/(9*x^2 + 1 2*x + 4) + 203/12*sqrt(-10*x^2 - x + 3)/(3*x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 324 vs. \(2 (90) = 180\).
Time = 0.39 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.66 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {607}{432} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {4}{135} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {77 \, {\left (157 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + 25480 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}\right )}}{18 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
607/432*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 4/135*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sqrt (2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 77/18*(157*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/ sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 25480*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqr t(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^3\,\sqrt {5\,x+3}} \,d x \]